12.2 Ratio test and integral test 607

17 With the aid of an appropriate figure show that, whens > 3,

3+4,+... +nd< (- Idx= 11c

2 x s-12 2

and hence that

and

Z,<c(S)-1<21,

lim ('(s) = 1.

(^18) Sketch the graph of f(x) = x 36 over the interval 0 < x < 1 and observe
that, even though f is unbounded and therefore not Riemann integrable, it can
be suspected that
(1) lim
()_4 I

r
n n 1 z 3` dx = 2,p
where the integral is a Cauchy extension of a Riemann integral. In any case,
use a result of Problem 14 to show that formula (1) is correct. Remark: One
who undertakes to prove (1) without use of Problem 14 does so at his own peril.
19 Suppose that f is nonnegative and continuous and increasing over the
interval x > 1 and that f(x) -> co as x --> co. Let

uk = f(k) (k = 1, 2, - ).

With the aid of a figure which is in some respects like Figure 12.24, show that to
each n there corresponds a number.4n such that 0 < In < I and

u1+U2+. .. +un= flnf(x) dx+nn-!In(un - u1).

Applying this to the case in which f (x) = log x and

n n
I uk = I log k = log n!,
k=1 k=1
obtain the formula

and hence

logn! =nlogn-n+l+(1 - .4.) log -n,

n! = nnl nenl-A* = (n/e)"enl-A..

Remark: This elementary calculation gives an introduction to the important idea
that n! is of the order of magnitude of (n/e)n. It is easy to see with the aid of
a figure that 4n is a little greater than ff' and hence that n1-An is less than.
In Problem 4 of Section 12.6, we shall discover (among other things) that if n
is a positive integer, then
n! = 2-. r nne- 0.112%
where Bn is a number for which

1 30,2< Bn < 1 30n2 + 105,4